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2bornot2b
How many members of the set S={x:1≤x≤100} are of the form 4n-1. I guess the answer is 25, but is there a way to prove it or show it?
Take the series of such numbers to be S. The nth term of that series will be given by 4n-1. Putting n=1,n=2 etc..you find that the series is: 3,7,11....... This is an arithmetic progression. 4n-1<100 4n<101 n<25.25 Therefore number of terms of the series would be 25.
i was trying some sigma sum notation but i couldn't get it to work Good answer @Mani_Jha
{3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, 71, 75, 79, 83, 87, 91, 95, 99} Here they are and they are 25. For proof, one proof above is right.